Turning the two solutions of a quadratic equation in a general solutionSo while I was working on a project for uni, I came across this quadratic equation: t 2 + 2 τ t − 1 = 0The solutions for this particular equation are: t 1 , 2 = τ ± τ 2 + 1 I also found somewhere on the internet that the solutions can be written as one single solution namely: t = s i g n ( τ ) | τ | + τ 2 + 1 But I didn't quite understand how they got to that conclusion, is this something that can be done with every quadratic equation with two solutions or is it only for this particular equation?Perhaps I should also mention that the equation always has a solution between -1 and 1 and that the solution with the sign function always gives the solution that is between -1 and 1.(The aforementioned equation is an equation that is used to compute givens-rotations, don't know if it is of any importance for the solution of the equation)

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Turning the two solutions of a quadratic equation in a general solutionSo while I was working on a project for uni, I came across this quadratic equation:      t    2    +  2  τ  t  −  1  =  0The solutions for this particular equation are:      t          1      ,      2        =  τ  ±            τ      2        +    1  I also found somewhere on the internet that the solutions can be written as one single solution namely:  t  =            s      i      g      n      (      τ      )                      |            τ              |            +                        τ          2                +        1            But I didn&#039;t quite understand how they got to that conclusion, is this something that can be done with every quadratic equation with two solutions or is it only for this particular equation?Perhaps I should also mention that the equation always has a solution between -1 and 1 and that the solution with the sign function always gives the solution that is between -1 and 1.(The aforementioned equation is an equation that is used to compute givens-rotations, don&#039;t know if it is of any importance for the solution of the equation)

Nigerkamg5 · Accepted Answer

Step 1First, let&#039;s check if your proposed solution makes sense. Let   τ  =  1, then       t          1      ,      2        =  1  ±      2  . If you are interested in the number with magnitude less than 1, then we are looking for   1  −      2    &amp;lt;  0but your proposed solution gives us a positive number. Hence it can&#039;t be the right formula.If   τ  ≥  0, then we have   τ  +            τ      2        +    1    ≥  1 and we have   −  1  ≤  τ  −            τ      2        +    1    &amp;lt;  0 by triangle inequality.Similarly, if   τ  ≤  0, we have   τ  −            τ      2        +    1   and   0  &amp;lt;  τ  +            τ      2        +    1    ≤  1 by triangle inequality as well.Step 2Hence when   τ  ≥  0, we want an expression of   τ  −            τ      2        +    1   and when   τ  ≤  0, we want an expression of   τ  +            τ      2        +    1  .That is we want an expression of   τ  −  s  i  g  n  (  τ  )            τ      2        +    1   with the convention that sign(0) can take value of 1 and -1.                    τ        −        s        i        g        n        (        τ        )                              τ            2                    +          1                                    =        (        τ        −        s        i        g        n        (        τ        )                              τ            2                    +          1                )        ⋅                              (            −            τ            −            s            i            g            n            (            τ            )                                          τ                2                            +              1                        )                                (            −            τ            −            s            i            g            n            (            τ            )                                          τ                2                            +              1                        )                                                            =                              −                          τ              2                        +            (                          τ              2                        +            1            )                                −            τ            −            s            i            g            n            (            τ            )                                          τ                2                            +              1                                                                        =                  1                      −            τ            −            s            i            g            n            (            τ            )                                          τ                2                            +              1                                                                        =                  1                      −            s            i            g            n            (            τ            )                          |                        τ                          |                        −            s            i            g            n            (            τ            )                                          τ                2                            +              1                                                                        =        −                              s            i            g            n            (            τ            )                                              |                        τ                          |                        +                                          τ                2                            +              1                                          The formula differs by a sign.

Turning the two solutions of a quadratic equation in a general solution. So while I was working on a project for uni, I came across this quadratic equation: t^2+2 tau t-1=0

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